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Rate Law MCQ - Practice Questions with Answers

Edited By admin | Updated on Sep 25, 2023 25:23 PM | #NEET

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  • Rate Law is considered one of the most asked concept.

  • 46 Questions around this concept.

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Consider the reaction, 2A+B\rightarrow products. When concentration of B alone was doubled, the half­-life did not change. When the concentration of A alone was doubled, the rate increased by two times. The unit of rate constant for this reaction is

The rate law for a reaction between the substances A and B  is given by rate =  k\left [ A \right ]^{n}\left [ B \right ]^{m}.

On doubling the concentration of A and halving the concentration of B, the ratio of the new rate to the earlier rate of the reaction will be as

In a reaction, A + B \rightarrow Product, the rate is doubled when the concentration of B is doubled, and the rate increases by a factor of 8 when the concentrations of both the reactants (A and B) are doubled, rate law for the reaction can be written as :

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For the reaction A+B\rightarrow products, it is observed that:

  1. On doubling the initial concentration of A only, the rate of reaction is also doubled and
  2. On doubling the initial concentrations of both A and B, there is a change by a factor of 8 in the rate of the reaction.

The rate of this reaction is given by:

 For the reaction,3A+2B\rightarrow C+D,  the differential rate law can be written as :

For a reaction between A and B, the order with respect to A is 2 and the order with respect to B is 3. The concentrations of both A and B are doubled, the rate will increase by a factor of:

Consider the reaction
\mathrm{3 A+2 B \longrightarrow 4 C }
\mathrm{\text { If }-\frac{d A}{d t}=k_{1}(c), \frac{-d B}{d t}=k_{2}(c), \frac{d C}{d t}=k_{3}(c)}
which of the following expression is true for \mathrm{k_{1}, k_{2}, k_{3}} ?

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Consider a reaction that follows the rate law \mathrm{R=K[A]^{3 / 2}}. If the concentration of \mathrm{A} is increased by \mathrm{100 \%}  by what percentage does the rate of the reaction change?

For a consecutive reaction, if the rate constant for the second step \mathrm{(B \rightarrow C)} is much Greater than the rate constant for the first step \mathrm{(A \rightarrow B)},what will be the overall rate expression for the formation of \mathrm{C} ?

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What is the rate-determining step of the reaction: \mathrm{2 \mathrm{NO}_2 \mathrm{Cl}^{\rightarrow} \rightarrow 2 \mathrm{NO}_2+\mathrm{Cl}_2}, if rate law of the chemical reaction is represented as: Rate \mathrm{=K\left[\mathrm{NO}_2 \mathrm{Cl}\right]}

Concepts Covered - 2

Rate Law

The rate of any reaction depends upon the concentration of the reactants in rate law equation that is rate law and rate of reaction depend upon the order of the reaction.
Let us consider this reaction:
\mathrm{xA}+\mathrm{yB} \rightarrow \text { Product }
Rate law equation for the reaction can be given as:
                            
                      \begin{array}{l}{\mathrm{R} \propto[\mathrm{A}]^{\mathrm{p}}[\mathrm{B}]^{\mathrm{q}}} \\\\ {\mathrm{R}=\mathrm{K}[\mathrm{A}]^{\mathrm{p}}[\mathrm{B}]^{\mathrm{q}}}\end{array}
\text { Order of reaction }=\mathrm{p}+\mathrm{q}
Here p, q are experimental quantities which may or may not be equal to the respective stoichiometric coefficients ( x, y).
 

Unit of Rate Constant

The differential rate expression for nth order reaction is as follows:
                       -\frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{k}(\mathrm{a}-\mathrm{x})^{\mathrm{n}}
\text { or } \quad \mathrm{k}=\frac{\mathrm{d} \mathrm{x}}{(\mathrm{a}-\mathrm{x})^{\mathrm{n}} \mathrm{dt}}=\frac{(\text { concentration })}{(\text { concentration })^{\mathrm{n}} \text { time }}=(\text { conc. })^{1-\mathrm{n}} \text { time }^{-1}
If concentration be expressed in mole L-1 and time in minutes, then
\mathrm{k}=\left(\text { mole } \mathrm{L}^{-1}\right)^{1-\mathrm{n}} \mathrm{min}^{-1}
\begin{array}{l}{\text { For zero order reaction, } \mathrm{n}=0 \text { and hence, } \mathrm{k}=\mathrm{mole} \mathrm{L}^{-1} \mathrm{min}^{-1}} \\ {\text { For first order reaction, } \mathrm{n}=1 \text { and hence, }} \\ {\: \mathrm{k}=\left(\text { mole } \mathrm{L}^{-1}\right)^{0} \min ^{-1}=\min ^{-1}} \\ {\text {For second order reaction, } \mathrm{n}=2 \text { and hence, }} \\ {\: \mathrm{k}=\left(\text { mole } \mathrm{L}^{-1}\right)^{-1} \min ^{-1}=\text { mole }^{-1} \mathrm{L} \min ^{-1}}\end{array}
The rate constant of a first-order reaction has only time in its unit. It has no concentration term in the unit. This means the numerical value of k first-order reaction is independent of the unit in which concentration is expressed. If the concentration unit is changed, the numerical value of k for a first-order reaction will not change. However, it would change with change in time.

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Rate Law
Unit of Rate Constant

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